A particle is projected from origin x-axis with velocity `V_0` such that it suffers retardation of magnitude given by `Kx^3` (where k is a positive constant).The stopping distance of particle is

Options

A. `((3V_0^2)/(4k))^(1/3)`
B. `((2V_0^2)/(k))^(1/4)`
C. `(V_0^2/(2k))^(1/4)`
D. `((2V_0^2)/(3k))^(1/3)`

Answer

To find the stopping distance of a particle projected from the origin along the x-axis with an initial velocity \( V_0 \) and experiencing a retardation of magnitude \( Kx^3 \), we can follow these steps:

### Step 1: Define the acceleration
The retardation (deceleration) is given as:
\[ a = -Kx^3 \]
This indicates that the acceleration is negative because the particle is slowing down.

### Step 2: Relate acceleration to velocity and displacement
Using the relationship between acceleration, velocity, and displacement, we have:
\[ a = \frac{dv}{dt} = v \frac{dv}{dx} \]
Thus, we can rewrite the equation as:
\[ -Kx^3 = v \frac{dv}{dx} \]

### Step 3: Rearranging the equation
Rearranging gives:
\[ -Kx^3 dx = v dv \]

### Step 4: Integrate both sides
Now, we will integrate both sides. The left side will be integrated from \( x = 0 \) to \( x = s \) (the stopping distance), and the right side will be integrated from \( v = V_0 \) to \( v = 0 \):
\[ \int_{0}^{s} -Kx^3 \, dx = \int_{V_0}^{0} v \, dv \]

### Step 5: Perform the integration
The left side integrates as follows:
\[ -\frac{K}{4} x^4 \bigg|_{0}^{s} = -\frac{K}{4} s^4 \]
The right side integrates to:
\[ \frac{v^2}{2} \bigg|_{V_0}^{0} = 0 - \frac{V_0^2}{2} = -\frac{V_0^2}{2} \]

### Step 6: Set the integrals equal
Setting the two sides equal gives:
\[ -\frac{K}{4} s^4 = -\frac{V_0^2}{2} \]

### Step 7: Solve for \( s^4 \)
Removing the negative signs and rearranging gives:
\[ \frac{K}{4} s^4 = \frac{V_0^2}{2} \]
Multiplying both sides by 4:
\[ K s^4 = 2 V_0^2 \]

### Step 8: Solve for \( s \)
Now, we solve for \( s \):
\[ s^4 = \frac{2 V_0^2}{K} \]
Taking the fourth root:
\[ s = \left( \frac{2 V_0^2}{K} \right)^{1/4} \]

### Step 9: Final expression for stopping distance
Thus, we can express the stopping distance \( s \) as:
\[ s = 2^{1/4} \cdot \left( \frac{V_0^2}{K} \right)^{1/4} = \frac{2^{1/4} V_0^{1/2}}{K^{1/4}} \]

### Step 10: Conclusion
The stopping distance of the particle is:
\[ s = \frac{2 V_0^2}{K^{1/4}} \]